Let $q=p^n$ where $p$ is a prime number. For which $q$ is the extensión $\mathbb{F}_{q^3}/\mathbb{F}_q$ (i.e the extension of degree $3$) an extension of the form $\mathbb{F}_q(\root 3 \of \alpha )$ where $\alpha \in \mathbb{F}_q$.
The only that I noticed is that char ($\mathbb{F}_q$)=3 ( $q=3^n$) , then since the frobenius automorphism is injective in a finite set, it's also surjective, so all the elements have a cubic root , therefore in characteristic 3, the extension of degree 3, are not "produced" by considering cubic roots. In the case of $char F \ne 3 $ I have no idea what to do.
If $q\equiv2\pmod3$, then every element of the field of $q$ elements is a cube, so the extension is not an extension by a cube root.
On the other hand, if $q\equiv1\pmod3$, then the field of $q$ elements has non-cubes in it, so the extension can be achieved by a cube root.
You have already addressed the case $q\equiv0\pmod3$.